I have this problem: Evaluate the integral (x^2-2)/(x^4+3) from 1 to infinity using a comparison test (evaluate if converging). I understand the basic idea is to find a function with a graph "above/below" this function that can be integrated. My solutions guide uses (3x^2)/(x^4) which of course is just 3x^-2. I see that it's easy to find the antiderivative of this and take it's limit at infinity. How did they come up with 3x^-2 as a comparison? Am I just missing some simple Algebra?
Improper integrals and comparison tests
Notice that the numerator is positive for all x > √2 and the denominator is positive for all real x.
In a fraction with both numerator and denominator being positive, increasing the numerator or decreasing the denominator will increase the value.
e.g. 2/3 > 2/5 because 3 < 5 and 3/5 > 2/5 because 3 > 2.
So using this when x > √2:
We can increase the numerator by 2 to x2, and decrease the denominator by 3 to x4.
We wind up with x2/ x4 > (x2-2) / (x4+3).
So we can use x2/x4 = x-2 as an upper bound for this interval. (3x-2 is usable too since it is even bigger than x-2).
The integrand is negative for 1<x<√2 so our bound applies in that region too and therefore is an upper bound for all x>1.
Thus ∫1∞ (x2-2)/(x4+3) dx < ∫1∞ x-2 dx = [-x-1]1∞ = 0 - (-1) = 1, which proves convergence.